You have found the following ages (in years) of all 6 bears at your local zoo: $ 14,\enspace 18,\enspace 11,\enspace 22,\enspace 2,\enspace 11$ What is the average age of the bears at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{14 + 18 + 11 + 22 + 2 + 11}{{6}} = {13\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $14$ years $1$ year $1$ year $^2$ $18$ years $5$ years $25$ years $^2$ $11$ years $-2$ years $4$ years $^2$ $22$ years $9$ years $81$ years $^2$ $2$ years $-11$ years $121$ years $^2$ $11$ years $-2$ years $4$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1} + {25} + {4} + {81} + {121} + {4}} {{6}} $ $ {\sigma^2} = \dfrac{{236}}{{6}} = {39.33\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{39.33\text{ years}^2}} = {6.3\text{ years}} $ The average bear at the zoo is 13 years old. There is a standard deviation of 6.3 years.